Without using truth tables prove that $$\neg(P\land Q)\to(\neg P\lor(\neg P\lor Q))\iff(\neg P\lor Q)$$
This is a question I've encountered during my examination.
So basically what I did was, I took the RHS of the equation which is $\neg(P\land Q)\to(\neg P\lor(\neg P\lor Q))$ and hoped I could equate it to $(\neg P\lor Q)$ even though I was sure it was not enough to prove this bi - conditional statement.
Following on my assumption, this is what I've reached
\begin{array}{rl} & & \neg(P\land Q)\to(\neg P\lor(\neg P\lor Q)) \iff\\ & \iff & (P\land Q)\lor(\neg P\lor Q) \iff\\ & \iff & (\neg P\lor Q\lor P) \land(\neg P\lor Q\lor Q) \iff\\ & \iff & Q \land(\neg P\lor Q) & \end{array}
I'm kinda stuck on what to do after this, I'm not even sure whether this approach is even right. So some help is appreciated on how to prove this bi - conditional statement.